A 25 kg crate starting from rest at the top slides down a plane that makes an angle of `30^(@)` with the horizontal . Whe it reaches the bottom of the 10 m long slide , its velocity is 8 m/s . The work done by the fore of friction is closest to a value of :

Options

A. `17xx10^(2)`J
B. `8.5 xx10^(2)`J
C. `6.5xx10^(2)`J
D. `4.5xx10^(2)`J

Explanation

To solve the problem, we will use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. Let's break down the steps to find the work done by the force of friction.

### Step 1: Identify the given values
- Mass of the crate (m) = 25 kg
- Length of the slide (L) = 10 m
- Angle of the incline (θ) = 30°
- Initial velocity (u) = 0 m/s (starting from rest)
- Final velocity (v) = 8 m/s

### Step 2: Calculate the height (h) of the incline
The height can be calculated using the sine function:
\[ h = L \cdot \sin(\theta) \]
Substituting the values:
\[ h = 10 \cdot \sin(30°) = 10 \cdot \frac{1}{2} = 5 \text{ m} \]

### Step 3: Calculate the gravitational potential energy (PE) at the top
The gravitational potential energy at the top of the incline is given by:
\[ PE = mgh \]
Substituting the values:
\[ PE = 25 \cdot 9.8 \cdot 5 = 1225 \text{ J} \]

### Step 4: Calculate the change in kinetic energy (KE)
The change in kinetic energy is given by:
\[ \Delta KE = KE_{final} - KE_{initial} \]
Where:
\[ KE_{final} = \frac{1}{2} mv^2 \quad \text{and} \quad KE_{initial} = 0 \text{ (since it starts from rest)} \]
Calculating the final kinetic energy:
\[ KE_{final} = \frac{1}{2} \cdot 25 \cdot (8)^2 = \frac{1}{2} \cdot 25 \cdot 64 = 800 \text{ J} \]
Thus, the change in kinetic energy is:
\[ \Delta KE = 800 - 0 = 800 \text{ J} \]

### Step 5: Apply the work-energy theorem
According to the work-energy theorem:
\[ W_{friction} + PE = \Delta KE \]
Rearranging gives:
\[ W_{friction} = \Delta KE - PE \]
Substituting the values:
\[ W_{friction} = 800 - 1225 = -425 \text{ J} \]

### Step 6: Conclusion
The work done by the force of friction is approximately \(-425 \text{ J}\).